Proof on Sequences: Sum of a convergent and divergent diverges

Prove if sequence [itex]a_{n}[/itex] converges and sequence [itex]b_{n}[/itex] diverges, then the sequence [itex]a_{n}[/itex]+[itex]b_{n}[/itex] also diverges.

2. Relevant equations

3. The attempt at a solution

My professor recommended a proof by contradiction. That is, suppose [itex]a_{n}[/itex]+[itex]b_{n}[/itex] does converge. Then, for every ε > 0, there exists a natural number [itex]N_{1}[/itex] so that n > [itex]N_{1}[/itex] implies |[itex]a_{n}[/itex]+[itex]b_{n}[/itex] - L|< ε

We already know there exists [itex]N_{2}[/itex] so that n > [itex]N_{2}[/itex] implies |[itex]a_{n}[/itex] - M| < ε. So let N = max{[itex]N_{1}[/itex], [itex]N_{2}[/itex]}. Then n > N means we know [itex]a_{n}[/itex] is "very close" to M. My purpose in this is to try and show that this implies [itex]b_{n}[/itex] has a limit (that is, it converges) providing a contradiction. However, I'm not sure how to go about this.

Proof: Assume [itex]a_{n}[/itex] is convergent and [itex]b_{n}[/itex] is divergent.

Now suppose that [itex]a_{n}[/itex]+[itex]b_{n}[/itex] is convergent.
Then [for every ε > 0 there exists a natural number [itex]N_{1}[/itex] so that
n > [itex]N_{1}[/itex] implies |[itex]a_{n}[/itex]+[itex]b_{n}[/itex] - L|< ε/2

We know by our assumption that there also exists natural number [itex]N_{2}[/itex] so that
n > [itex]N_{2}[/itex] implies |[itex]a_{n}[/itex]-M| < ε/2